![]() My assumption is, the raid-controller takes 1 parity bit (for example disk 1's parity bit) and after that, from the 5th blocks of disks (5,6,7,8th block of 2,3rd disk), it XOR's the parity bit of the 1st disk's parity(4th block of 1st disk (parity)) and the data's in 5,6,7,8th blocks of disk 2 and 3. How is the data in the 5,6,7 and 8th block in disk 4 will be recovered? In this scenario, the RAID 5's explanation is accurate. The correct picture should look like this: Disk 1 | Disk 2 | Disk 3 | Disk 4 But, the RAID 5 explanations doesn't explain it like that and also there is RAID 6 configurations whose speciality is to have 2 parity bit. ![]() , so that there is 2 parity information in 1 disk (Parity information should be more if there is more disks). However, in order to recover data from 8 blocked storage for example, we need something like Figure b. The thing is, as I mentioned, all the sources says that RAID 5 has 1 parity bit in every disk as I showed in figure a. This makes sense in 4 disk 4 block scenario because if for example disk 4 fails, we can recover all of the data's because we got the exact amount of parity bits to recover all the data that is lost in disk 4.īut, in order to recover the data in more blocked disks (for example 8 blocks), the picture should be like this: Disk 1 | Disk 2 | Disk 3 | Disk 4 Internet sources that demonstrate RAID 5 shows that, every disk has a Parity bit (only 1) in their blocks. ![]() And in every demonstration, all sources shows raid 5 like: Disk 1 | Disk 2 | Disk 3 | Disk 4 So that, in every explanation of RAID 5, it is said that RAID 5 has parity bit in every disk. This is an old topic but I have a question about RAID 5 parity bits that confuse me.
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